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A problem on triangle and it's perpendicular bisectors.

  1. Aug 17, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm trying to solve the following problem :

    In △ABC, coordinates of B are (−3,3). Equation of the perpendicular bisector of side AB is 2x+y−7=0. Equation of the perpendicular bisector of side BC is 3x−y−3=0. Mid point of side AC is E(11/2,7/2). Find AC2.


    2. Relevant equations

    -

    3. The attempt at a solution

    Fw5PT.png

    By solving 3x−y−3=0 and 2x+y−7=0 I find that the intersection of perpendicular bisectors is at (2,3) .
    Then using the two points (2,3) and (11/2,7/2), I get the equation of perpendicular bisector of AC as y=x/7+19/7.
    So the slope of AC is -7 and then using point slope form , y−7/2=−7(x−11/2) Thus the equation of line AC is y=42−7x .
    Similarly equation of line BC is y=2−x/3 .
    So AC and BC intersect at (6,0).
    By using the fact that E is the midpoint of AC, I find Co-ordinates of A as (5,7).
    So the distance between A and C is 5√2, and AC2=50.
    But this answer is wrong and the correct answer is 74 ( I checked the answer sheet) .
    What have I done wrong ?
     
  2. jcsd
  3. Aug 17, 2013 #2

    verty

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    I don't know what went wrong before but here is a slightly different method.

    Step 1. Find A using B and the gradient of AB.
    Step 2. Find AC^2

    I forget about E, E means only A is required.
     
  4. Aug 26, 2013 #3
    I don't understand. How can I find AC2 after finding gradient of AB ?
     
  5. Aug 26, 2013 #4

    verty

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    Your goal is to find A and then AC^2. To find A, find the midpoint of AB then use the midpoint formula.
     
  6. Aug 26, 2013 #5

    haruspex

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    I agree with your answer, through a slightly different route.
    O is the circumcentre, so OC=OB = 5. OE2 = 25/2, so CE2 = 25/2 by Pythagoras.
     
  7. Aug 27, 2013 #6
    I tried and got A as (5,7) . Is that right?
    Then I get C as (6,0) so AC2=50.
     
  8. Aug 28, 2013 #7

    verty

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    It must be right, you found the same answer by two different methods and Haruspex found it by a third method. There must have been no mistake originally.
     
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